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NVAEasyJEE 2024Diffraction & Polarisation

JEE Physics 2024 Question with Solution

A parallel beam of monochromatic light of wavelength 5000 A˚5000 \, \text{Å} is incident normally on a single narrow slit of width 0.001 mm0.001 \, \text{mm}. The light is focused by a convex lens on a screen, placed on its focal plane. The first minima will be formed for the angle of diffraction of:

Answer

Correct answer:30

Step-by-step solution

Standard Method

Given: wavelength λ=5000 A˚=5000×10−10 m\lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m}, slit width a=0.001 mm=1×10−6 ma = 0.001 \, \text{mm} = 1 \times 10^{-6} \, \text{m}.

Find: the angle of diffraction θ\theta for the first minima.

For the first minima in single-slit diffraction,

asin⁡θ=λa \sin \theta = \lambda

Substituting the given values,

1×10−6sin⁡θ=5000×10−101 \times 10^{-6} \sin \theta = 5000 \times 10^{-10}

So,

sin⁡θ=5000×10−101×10−6=0.5\sin \theta = \frac{5000 \times 10^{-10}}{1 \times 10^{-6}} = 0.5

Hence,

θ=sin⁡−1(0.5)=30∘\theta = \sin^{-1}(0.5) = 30^\circ

Therefore, the angle of diffraction for the first minima is 30∘30^\circ.

Direct Ratio Method

Given: a=1×10−6 ma = 1 \times 10^{-6} \, \text{m} and λ=5000×10−10 m\lambda = 5000 \times 10^{-10} \, \text{m}.

Find: the first diffraction minima angle.

For first minima, directly use

sin⁡θ=λa\sin \theta = \frac{\lambda}{a}

Thus,

sin⁡θ=5000×10−101×10−6=0.5\sin \theta = \frac{5000 \times 10^{-10}}{1 \times 10^{-6}} = 0.5

So,

θ=30∘\theta = 30^\circ

This works because the first minima corresponds to order m=1m = 1 in the condition asin⁡θ=mλa \sin \theta = m\lambda. Therefore, the required angle is 30∘30^\circ.

Common mistakes

  • Using the interference formula instead of the single-slit diffraction condition is incorrect because this question asks for the first minima of a single narrow slit. Use asin⁡θ=mλa \sin \theta = m\lambda, not a double-slit relation.

  • Not converting units properly gives a wrong value of sin⁡θ\sin \theta. Convert 0.001 mm0.001 \, \text{mm} to 1×10−6 m1 \times 10^{-6} \, \text{m} and 5000 A˚5000 \, \text{Å} to 5000×10−10 m5000 \times 10^{-10} \, \text{m} before substitution.

  • Taking the wrong order of minima is a conceptual error. For the first minima, use m=1m = 1, not m=2m = 2 or higher. That is why the condition becomes asin⁡θ=λa \sin \theta = \lambda.

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